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A body moving with constant acceleration | travels the distance 3 metre and 8 metre respectively in 1 second and 2 second | calculate the initial velocity and the acceleration of body | Sum Motion in one Dimension - Physics class 9

Problem :     A body moving with constant acceleration travels the distance 3 metre and 8 metre respectively in one second 2 second. Calculate the 1) initial velocity and 2) the acceleration of body. Data given:
Distance(s1) = 3m
Time taken (t1) to travel 3 m =1 second
Distance(s2) = 8m
Time taken (t2) to travel 8 m =2 second
Initial velocity (u)= ?
Acceleration (a) = ?
Case 1:We know
s = ut + 1/2 a. t2
Or s1 = ut1 + 1/2 a. t12
or 3 = u . 1 + 1/2 .a . 12
or 3 = u + a/2 -------Eqn. 1 Case 2: s2 = ut2 + 1/2 a. t22
or 8 = u . 2 + 1/2 .a . 22
or 8 = 2u + 2a
or 8/2 = (2u + 2a)/2
or 4 = u + a -------Eqn. 2
Subracting Eqn. 1 from Eqn. 2 (Eqn. 2 - Eqn. 1) We have:
4 - 3 = (u + a) - (u + a/2)
Or 1 = u + a - u - a/2
Or 1 = a - a/2
Or 1 = (2a - a)/2
Or 1 = a /2
Or a = 2 m/s2
Putting the value of a in Eqn. 1 we have:
3 = u + 2/2
3 = u + 1
Or u = 3 - 1
Or u = 2 m/s
∴ Initial velocity (u)= 2 m/s
Acceleration (a) = 2 m/s2Answer

Summary of Journey by Night | For Classv9

Sher Singh was a 12 year old brave boy; who lived in Laldwani village. He lived with his younger brother, mother and father. Sher Singh Bahadur was father of Sher Singh. Sher Sing's younger brother's name was Kunwar. He was a farmer by occupation but he was a famous hunter too. Whenever there was an expedition in the village, people used to send him (Sher Singh Bahaur). He was an experienced man at hunting. When Sher Singh's brother was seriously ill, his father had gone for a photography expedition with other villagers of Laldwani as beaters. His mother had experience of controlling capital for carrying load and it was their also earning source. They were very much poor, therefore his mother was unable to leave her job to bring her ailing son to a doctor. Actually Kunwar was suffering from influenza. In Laldwani village diseases Cholera and Influenza had taken many lives . Kunwar had a bad stomach ache and his body temperature was more than normal. Sher Sin…

Walking at 3/5 of his usual speed, a man is 4 hours too Late | Find his usual time | Speed Distance and Time

Walking at 3/5 of his usual speed,  a man is 4 hours too late.  Find his usual time. Suppose the usual speed of the man be x km/hr .
Let distance travelled be D km
Therefore usual Time taken = D / x
.'. walking speed = 3/5. x
Let distance travelled be D km
we can write
D/(3/5 . x)- D/x = 4
Or D((5 / 3. x) - 1/x) = 4
Or D/x ((5 / 3 ) - 1) = 4
Or D/x (2 / 3) = 4
But D/x = Usual Time taken by the man .
T= 12/2
or T= 6 hours . Answer

A body moves from rest with uniform acceleration and travels 270 metre in 3 seconds | Find the velocity of the body at 10 seconds after the start | Sum Motion in one Dimension - Physics class 9

Problem : A body moves from rest with uniform acceleration and travels 270 metre in 3 seconds. Find the velocity of the body at 10 seconds after the start. Initial velocity (u)= 0
Distance(s) = 270m
Time taken (t)=3s
final velocity after 10 s (v10)= ?
First we will calculate acceleration.
We know
s = ut + 1/2 a. t2
or 270 = 0 . 3 + 1/2 . a. (3)2
Or a= (270 . 2) / 9
Or a= 60 m/s2
Again We know that:
a = (v - u ) / t
Or 60 = (v - 0 )/ 10
Or v = 60 . 10
∴ v = 600 m/sAnswer

When brakes are applied to a bus the retardation produced is 25 cm/s2 and the bus takes 20 s to stop | calculate the initial velocity of bus the distance traveled by bus during this time |

Problem : When brakes are applied to a bus the retardation produced is 25 centimetre per second square (cm/s2 )and the bus takes 20 seconds to stop. Calculate a)the initial velocity of bus b) the distance traveled by bus during this time Data given:
Retardation (-a) = - 25 cm/s2
Time taken to get stopped (tav)=20 second
As the bus starts from rest,
∴ final velocity (v)= 0
Initial velocity (u)= ?
Distance(s) = ?
First we will calculate initial velocity of the bus.
We know
a = (v - u ) / t
Or - 25 = (0 - u )/ 20
Or -25 = - u /20
Or u = 25x 20
∴ u = 500 cm/s
Or u = (500/100) m/ s
Or u= 5 m/s

Again We know that:
s = ut + 1/2 a. t2
or S = 500x 20 + 1/2 .(-25). (20)2
or S = 10000 - 1/2 .25. 400
or S = 10000 - 5000
or S = 5000 cm
or S = 50 m
∴ Initial velocity (u) = 5m/s and
Distance (s)=50m Answer

A car travels a distance 100 metre with a constant acceleration and average velocity of 20 metre per second | The final velocity required by the car is 25 metre per second | Find the initial velocity and acceleration of car | Sum Motion in one Dimension - Physics class 9

Problem: A car travels a distance 100 metre with a constant acceleration and average velocity of 20 metre per second. The final velocity required by the car is 25 metre per second. Find 1) the initial velocity and 2) acceleration of car Data given:
Distance(s) = 100 metre
Acceleration(a) = ? ( to calculate)
Average velocity(vav)=20 metre per second
final velocity (v)= 25 metre per second
First we will calculate time taken to complete the 100 metre distance with average velocity of 20 metre per second.
t = s/vav
Or t = 100/20
∴ t = 5 s


We know:
a = (v-u)/t
a = (25-u)/5

again we know that:
v2 = u2+2as
Or (25)2=u2+2.a.100
Putting the value of a we have:
Or 625 = u2+200. ((25-u)/5)
Or 625 = u2+1000-40u
Or u2-40u+1000-62 5= 0
Or u2-40u+375 = 0
Or u2-25u-15u+375 = 0
Or u (u-25)-15(u-25)= 0
Or (u-25)(u-15)=0
∴ u=25 u=15 Now when u=25
a=(25-25)/5
Or a=0
Which is absurd as the body is moving with constant acceleration. when u=15
a= (25-15)/5
∴ a= 2 m/s2
∴ u = 5 m/s
And a=2 m/s…